Sum of Diagonal Entries of Matrix M

Let $M$ be a $3 \times 3$ matrix represented by its column vectors $C_1, C_2, \text{ and } C_3$:

$$M = \begin{bmatrix} C_1 & C_2 & C_3 \end{bmatrix} = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$$

From the first given equation:

$$M \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = C_2 = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix}$$

Thus, the second column of $M$ is:

$$C_2 = \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} \implies b = -1, \quad e = 2, \quad h = 3$$

From the second given equation:

$$M \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = C_1 - C_2 = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}$$

Substituting the value of $C_2$, we find $C_1$:

$$C_1 = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} + C_2 = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} + \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 2 \end{bmatrix}$$

Thus, the first column of $M$ is:

$$C_1 = \begin{bmatrix} 0 \\ 3 \\ 2 \end{bmatrix} \implies a = 0, \quad d = 3, \quad g = 2$$

From the third given equation:

$$M \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = C_1 + C_2 + C_3 = \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix}$$

Substituting $C_1$ and $C_2$, we find $C_3$:

$$C_3 = \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix} - C_1 - C_2 = \begin{bmatrix} 0 \\ 0 \\ 12 \end{bmatrix} - \begin{bmatrix} 0 \\ 3 \\ 2 \end{bmatrix} - \begin{bmatrix} -1 \\ 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 1 \\ -5 \\ 7 \end{bmatrix}$$

Thus, the third column of $M$ is:

$$C_3 = \begin{bmatrix} 1 \\ -5 \\ 7 \end{bmatrix} \implies c = 1, \quad f = -5, \quad i = 7$$

Hence, the full matrix $M$ is:

$$M = \begin{bmatrix} 0 & -1 & 1 \\ 3 & 2 & -5 \\ 2 & 3 & 7 \end{bmatrix}$$

The diagonal entries of $M$ are $a = 0$, $e = 2$, and $i = 7$.

The sum of the diagonal entries (trace of $M$) is:
$\text{Trace}(M) = a + e + i = 0 + 2 + 7 = 9$